I had already worked this problem out for someone else using pills. The setting of the problem, as I remember it, goes something like this.

A wise man has committed a capital crime and is to be put to death. The king decides to see how wise the man is. He gives to the man 12 pills which are identical in size, shape, color, smell, etc. However, they are not all exactly the same. One of them has a different weight. The wise man is presented with a balance and informed that all the pills are deadly poison except for the one which has a different weight. The wise man can make exactly three weighings and then must take the pill of his choice.

Remember that the only safe pill has a different weight, but the wise man doesn't know whether it is heavier or lighter.

Most people seem to think that the thing to do is weigh six pills against six pills, but if you think about it, this would yield you no information concerning the whereabouts of the only safe pill.

So that the following plan can be followed, let us number the pills from 1 to 12. For the first weighing let us put on the left pan pills1,2,3,4 and on the right pan pills 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the good pill is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the good pill is either 11 or 12.

Weigh pill 1 against 11. If they balance, the good pill is number 12. If they do not balance, then 11 is the good pill.

If 1,2 vs 9,10 do not balance, then the good pill is either 9 or 10. Again, weigh 1 against 9. If they balance, the good pill is number 10, otherwise it is number 9.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these pills could be the safe pill. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the good pill is either 3 or 4. Weigh 4 against 9, a known bad pill. If they balance then the good pill is 3, otherwise it is 4.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a good, heavy pill, or 1 is a good, light pill.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the good pill. If they balance, then 1 is the good pill. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a good heavy pill or 2 is a light good pill. Weigh 5 against 6. The heavier one is the good pill. If they balance, then 2 is a good light pill.

I think that one could write all of this out in a nice flow chart, but I'm not sure that the flow chart would show up correctly over e-mail.

Santiago's doesn't work because of the heavier or lighter, you can't guarantee that you'll make a correct choice when you pick between the first groups. It could be a heavier person in the side weighing down the seesaw or a lighter person in the other group.

Dark Dragon 64 wrote:I still think that Santiago was right. How does your method work in three attempts?

Santiago's works if one of them's heavier - but one of them is either heavier or lighter.

Separate them into 3 groups of 4 people

Group AGroup BGroup C

Weigh group A and group B

The results of this then tell you which way to go

Scenario 1: Groups A and B are equal

You know Group C contains the Heavy/Lighter

Test any random from group C against any 3 from groups A&B (as you know they are all the same)

If these 2 groups of 3 are equal, you know the odd one is the remaining from group C. Test him against anyone else to work out if he's heavier or lighter.

If these 2 groups of 3 are not equal, then you now know if the odd one out is heavier or lighter, and that he's in the group of 3 from C. Weigh 2 of of 3 from Group C and then you can determine which he is

Scenario 1: Groups A and B are not equal

This gives you 4 possible lighters (L) and 4 possible heavies (H) - depending on which of A/B went up/downYou also have 4 "Normals" from Group C (N)

Create 3 groups as follows1 - LLH2 - HHL3 - HLK

Now test group 1 against group 2, to get 3 possible outcomes

Outcome 1 - they're equal, then group 2 contains the odd one out

Weigh the two Hs (remembering they're only possibly heavier) from group 2 against each other. Either one of them will be heavier, of if their equal you know the L is the lighter one.

Outcome 2 - Group 3 is heavier then you know that the odd one out is either the H in group 3 is the odd one out or one of the 2 Ls from group 1 is the odd one out.Wigh the two Ls against each other to see which of these 3 is the odd man out.

Outcome 3 - Group C is lighter, then either the L from group 3 is the odd one out or the H from group 1 is. Simply weigh either one against other

I watched the last two episodes together and they made for an excellent end to the season.

It's a shame to see Holt go and I hope that we haven't seen the last of him. Terry and Gina made a great team as they worked to save Holt as Terry lifted the filing cabinet . I felt that Amy and Jake's relationship was forced to develop too quickly in the last episode but I liked Diaz's reaction to her birthday 'party' and Charles' behaviour throughout it all.